Science:Math Exam Resources/Courses/MATH307/December 2012/Question 03 (b)/Solution 1

If S is the null space of the matrix A from part (a), then the basis for S can be found as the basis of N(A). Let's choose the simplest form $A={\begin{bmatrix}1&1&1\end{bmatrix}}.$

We see that we choose x₂ and x₃ to be free variables when we set x₁ = -x₂-x₃. That is, the kernel N(A) (and with it the subspace S) can be expressed as

${\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}}=x_{2}{\begin{bmatrix}-1\\1\\0\end{bmatrix}}+x_{3}{\begin{bmatrix}-1\\0\\1\end{bmatrix}}$

In other words, a basis for S is given by the two vectors

$b_{1}={\begin{bmatrix}-1\\1\\0\end{bmatrix}},\quad b_{2}={\begin{bmatrix}-1\\0\\1\end{bmatrix}}$